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If a ball is thrown upward at 64 feet per second from a height of 8 feet, the height of the ball can be modeled by [tex]S = 8 + 64t - 16t^2[/tex] feet, where [tex]t[/tex] is the number of seconds after the ball is thrown. How long after the ball is thrown is the height 68 feet?

It takes ___ seconds for the ball to reach the height of 68 feet.

Answer :

The ball reaches a height of 68 feet after 1.25 seconds.

To determine the time it takes for the ball to reach a height of 68 feet, we can set the equation S = 68 and solve for t. Substituting the values into the equation S = 8 + 64t - 16t^2, we get 68 = 8 + 64t - 16t^2. Rearranging the equation and setting it equal to zero, we have 16t^2 - 64t + 60 = 0. Factoring this quadratic equation, we get (4t - 6)(4t - 10) = 0. Solving for t, we find two possible values: t = 1.5 and t = 2.5. However, since we are considering the upward motion of the ball, we take the positive value, which is t = 1.25 seconds.

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